Sodium ions (Na+) move at 0.844 m/s through a bloodstream in the arm of a person standing near a large magnet. The magnetic field has a strength of 0.260 T and makes an angle of 50.0° with the motion of the sodium ions. The arm contains 110 cm3 of blood with a concentration of 2.00 1020 Na+ ions per cubic centimeter. If no other ions were present in the arm, what would be the magnetic force on the arm?

Respuesta :

Answer:

F = 591.8 N

Explanation:

Force on a moving charge in constant magnetic field is given as

[tex]F = q(\vec v \times \vec B)[/tex]

so we can say

[tex]F = qvBsin\theta[/tex]

here we know

[tex]q = 1.6 \times 10^{-19} C[/tex]

[tex]v = 0.844 m/s[/tex]

[tex]B = 0.260 T[/tex]

[tex]\theta = 50 degree[/tex]

now we have

[tex]F = (1.6 \times 10^{-19})(0.844)(0.260)sin50[/tex]

[tex]F = 2.69 \times 10^{-20} N[/tex]

Now concentration of Sodium ions is given as

[tex]C = 2.00 \times 10^{20} ions/cm^3[/tex]

total volume of blood = 110 cc

now total number of ions is given as

[tex]N = (2 \times 10^{20})(110) = 2.2 \times 10^{22} ions[/tex]

now total force on the hand is given as

[tex]F = (2.69 \times 10^{-20})(2.2 \times 10^{22})[/tex]

[tex]F = 591.8 N[/tex]

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